Question: Graph this system of equations and solve. $8x+10y = -10$ $y = -2 x + 5$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Explanation: Convert the first equation, $8x+10y = -10$ , to slope-intercept form. $y = -\dfrac{4}{5} x - 1$ The y-intercept for the first equation is $-1$ , so the first line must pass through the point $(0, -1)$ The slope for the first equation is $-\dfrac{4}{5}$ . Remember that the slope tells you rise over run. So in this case for every $4$ positions you move down (because it's negative) You must also move $5$ positions to the right. $5$ positions to the right. $4$ positions down from $(0, -1)$ is $(5, -5)$ Graph the blue line so it passes through $(0, -1)$ and $(5, -5)$ The y-intercept for the second equation is $5$ , so the second line must pass through the point $(0, 5)$ The slope for the second equation is $-2$ . Remember that the slope tells you rise over run. So in this case for every $2$ positions you move down (because it's negative) You must also move $1$ positions to the right. $1$ position to the right. $2$ positions down from $(0, 5)$ is $(1, 3)$ Graph the green line so it passes through $(0, 5)$ and $(1, 3)$ The solution is the point where the two lines intersect. The lines intersect at $(5, -5)$.